Problem: Is ${220569}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {220569}= &&{2}\cdot100000+ \\&&{2}\cdot10000+ \\&&{0}\cdot1000+ \\&&{5}\cdot100+ \\&&{6}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {220569}= &&{2}(99999+1)+ \\&&{2}(9999+1)+ \\&&{0}(999+1)+ \\&&{5}(99+1)+ \\&&{6}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {220569}= &&\gray{2\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {2}+{2}+{0}+{5}+{6}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${220569}$ is divisible by $3$ if ${ 2}+{2}+{0}+{5}+{6}+{9}$ is divisible by $3$ Add the digits of ${220569}$ $ {2}+{2}+{0}+{5}+{6}+{9} = {24} $ If ${24}$ is divisible by $3$ , then ${220569}$ must also be divisible by $3$ ${24}$ is divisible by $3$, therefore ${220569}$ must also be divisible by $3$.